Part 2: determine the number of protons, electrons, and neutrons present in an atom of potassium (k). explain how you determined your answer using complete sentences.

Well first we can look at the periodic table and understand what the numbers mean. 
First we look at the top right of the element (AkA) The atomic number which indicates how many protons are in the single nucleus of a single potassium atom, In this case its 19.
Then there Is another number under neath the element that is used to calculate the number of neutron's that are in a single potassium atom. This number is called the atomic weight, In this case the number is 39.098.
First the number is rounded and since the number is in the tenths place we round it downs (No larger than 5) to 39 then we subtract the atomic number to get the typical amount of neutrons in a typical potassium atom: 20

Hope it helps :)

STEPS:

1) NUMBER OF PROTONS: The number of protons is always the same as the element's atomic number. So, since Potassium's atomic number is 19, Potassium has 19 protons.

2) NUMBER OF ELECTRONS: In neutral atoms, the number of electrons is always the same as the number of protons. So (unless it was specified that this atom is an ion), Potassium has 19 electrons.

3) NUMBER OF NEUTRONS: The number of neutrons is the element's mass number (rounded from the average atomic mass) minus the number of protons. So do the math:
               39-19 = 20
So Potassium has 20 neutrons.

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.